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a skydiver steps out of an airplane. the velocity-time graph shows how …

Question

a skydiver steps out of an airplane. the velocity-time graph shows how her velocity changes with time. (note: down is defined as the + direction of velocity.) at what points during the skydive does she experience the greatest, a zero, and the smallest (but non - zero) air resistance?

Explanation:

Step1: Recall air resistance and acceleration relation

Air resistance \( F_{air} \) opposes motion. The net force \( F_{net}=mg - F_{air} \) (downward is +, \( mg \) is weight, \( F_{air} \) is upward). Acceleration \( a=\frac{F_{net}}{m}=g-\frac{F_{air}}{m} \). On a velocity - time graph, the slope is acceleration (\( a = \frac{\Delta v}{\Delta t} \)).

Step2: Analyze greatest air resistance

Greatest \( F_{air} \) means smallest acceleration (since \( a = g-\frac{F_{air}}{m} \), larger \( F_{air} \) gives smaller \( a \)). The slope of the velocity - time graph is smallest (closest to zero) at point D (where the graph is flat, slope = 0 at D? Wait, no, D is when velocity is constant, so acceleration \( a = 0 \). Wait, no: initially, at A, slope is steep (large acceleration). As time increases, slope decreases. At D, velocity is constant, so acceleration \( a = 0 \). So \( a = g-\frac{F_{air}}{m}=0\Rightarrow F_{air}=mg \), which is the greatest air resistance (since it balances weight, largest \( F_{air} \)).

Step3: Analyze zero air resistance

Zero \( F_{air} \) means \( a = g \) (since \( a = g-\frac{0}{m}=g \)). The slope of the velocity - time graph is largest (steepest) at point A (initial point, just after jumping, before air resistance builds up much). At A, the slope is maximum, so \( a = g \), meaning \( F_{air}=0 \) (since \( a = g-\frac{F_{air}}{m}=g\Rightarrow F_{air}=0 \)).

Step4: Analyze smallest non - zero air resistance

Smallest non - zero \( F_{air} \) means acceleration is slightly less than \( g \), so the slope is slightly less than the slope at A. The point with the next steepest slope (after A) is B? Wait, no: the slope (acceleration) decreases as \( F_{air} \) increases. So the smallest non - zero \( F_{air} \) occurs where acceleration is just slightly less than \( g \), which is at point B? Wait, no, let's re - order:

  • Zero air resistance: At A (slope is maximum, \( a = g \), so \( F_{air}=0 \)).
  • Smallest non - zero air resistance: At B (slope is less than at A, so \( a = g-\frac{F_{air}}{m} \) is less than \( g \), so \( F_{air} \) is non - zero and small).
  • Greatest air resistance: At D (slope is zero, \( a = 0 \), so \( F_{air}=mg \), greatest \( F_{air} \)).

Answer:

  • Greatest air resistance: Point D (where velocity is constant, acceleration \( a = 0 \), so \( F_{air}=mg \), maximum air resistance).
  • Zero air resistance: Point A (initial point, slope is maximum, acceleration \( a = g \), so \( F_{air}=0 \)).
  • Smallest non - zero air resistance: Point B (slope is less than at A, acceleration is less than \( g \), so \( F_{air} \) is non - zero and small).